User blog:Edwin Shade/Is Pi ''Really'' Transcendental ?
If you were to answer the question that the title of this blog asks, you may say quickly: "Yes ! Pi is transcendental ! After all, Ferdinand von Lindemannhttps://en.wikipedia.org/wiki/Ferdinand_von_Lindemann proved that pi cannot ''be expressed as the root of ''any ''polynomial equation with integer coefficients and exponents in 1882." This is true. Yet as we are about to learn, it may be that mathematicians have yet to ''truly prove the transcendentalism of any known number. To begin with, let us rehash what it means for a number to be transcendental. In order for a number to be transcendental it means there exists no polynomial equations with integer coefficients and exponents such that one of it's roots is that number. \(\frac{3}{7}\) is not transcendental because it is the root of the linear equation \(7x-3=0\), and likewise the cube root of seven is not transcendental because it is the root to the equation \(x^3-7=0\). It seems then that if a number is proved to be transcendental then there is no function it is a root of. This though, is untrue because the root of the equation \(x^x-2=0\) may be transcendental, yet it is describable as the solution to the equation \(x^x-2=0\). You see, the definition for transcendentality only takes into consideration whether polynomial equations can describe the number, when in fact there are an infinity of functions beyond polynomial's, including tetrational functions for instance, like \(^4x+5x^x-7=0\). Now I will define a set of numbers that are truly ''transcendental, that is, they are not expressible as the solutions of ''any function ''composed entirely of integers and hyper-operators. First let us create a set of numbers, which I will call "Class-0" numbers. This set of numbers is defined as, "the set of all numbers that are the solutions to equations which are composed of terms that contain at most the 0th hyper-operator, (that is, successorship)." Clearly, only one equation can be written out that uses ''at most ''the 0th hyper-operator. That equation is \(x+1=0\), and there is only one solution, namely, \(x=-1\). Class-0 numbers then consist of a set of numbers with only one element, -1. Next let us create an additional set of numbers, which I will call "Class-1" numbers. This set of numbers can be defined as, "the set of all numbers that are the solutions to equations which are composed of terms that contain at most the 1st hyper-operator, (that is, addition)." There are now ''infinitely many more equations that contain at most only the 1st hyper-operator, and they are all of the form \(x+a=0\). It is obvious then that Class-1 numbers are the set of all integers, because any integer may be described as the solution to an equation of the form \(x+a=0\). Now let us generalize this concept of "number classes" by stating the nth-Class of numbers are "the set of all numbers that are the solutions to equations which are composed of terms that contain at most the Nth-hyper-operator." More formally, Class-N numbers contain the set \(\{x:\exists F_N(x)=0\}\), Where \(F_N(x)\) is the set of all functions composed of terms which contain at most the Nth hyper-operator. Class-0 numbers consist of only a single element, -1; Class-1 numbers consist of the set of all integers; Class-2 numbers consist of the set of all numbers expressible as fractions, (all rational numbers); and Class-3 numbers include all polynomial roots real as well as complex. Is it Class-4 numbers that are of interest to us, because if a number is a Class-4, then is it considered by Mathematicians to be "transcendental", even though it is expressible as the solution to an equation with tetrational operations. Let the set of all numbers that can be described as the solution to such equations be called the "N-Class" of numbers. So then, for a number to be transcendental in the true sense of the word it must fall outside ''of the set of ''all ''number classes, or belong to the set \(\{x:\lnot \exists F_N(x)=0\}\), which I will call the TT-Class of numbers. (TT stands for "Truly Transcendental"). Although the of all number classes seems to fill the entire number line, there a simple proof for why there will always be "unreachable" numbers that fall into the TT-Class. As Cantor demonstrated, you can enumerate a countable collection of countable sets, and thus enumerate the set of all rational numbershttps://www.youtube.com/watch?v=elvOZm0d4H0,(Godel proved this). This however can be extended to include more than just two sets of integers, and can be used to count all the different functions possible with N-order hyper-operators and an arbitrarily high number of terms. By mapping each function to a point in higher-dimensional space these functions can be enumerated much the same way that the set of all fractions can be enumerated. As a natural consequence then, the set of all solutions of these functions, (and hence all Class-N numbers), can be enumerated. Even disregarding the complex solutions, this would mean that if Class-N numbers filled the real number line completely and yet were still enumerable then the set of all real numbers would be countable. The set of real numbers was proven to be ''uncountable though, and so it contradicts our assertion that there are no numbers outside of the set of Class-N numbers.. Thus, there must exist a TT-Class of numbers, as defined above, which contains the set of all numbers not describable as N-Class numbers. In conclusion, it can be seen that although it has been proven that pi is ''not a member ''of the Class-3 set of numbers, there appears to be no proof that pi is a member of the TT-Class of numbers. It could just as well be a member of the Class-179,297 set of numbers, and none would be the wiser till it was proven. After all, to prove a number does not belong to the Nth-Class of numbers seems to get progressively more difiicult, it took hundreds of years before someone proved the irrationality of \(\sqrt{2}\), and then almost two thousands before someone proved Pi is "transcendental". In order for it to be proved a number belongs to the TT-Class of numbers it would have to proven it is not a member of any N-Class number set. Thus my question is this, how may it be proven within a finite space and time that any number definable in such a way that allows us to calculate it's digits to arbitrary precision is either N or TT-Class ? Your thoughts on this are welcome below, for I am well aware the question I just asked may go unresolved for a while, but it is just something to think about. Sources Category:Blog posts